Recall that \(m\) , \(s\)  and \(t\)  are Riemannian submersions. Thus, the vector fields \(\nabla (f\circ s)\)  and \(\nabla f\)  are \(s\) -related, \(\nabla (f\circ t)=\nabla (f\circ s)\)  and \(\nabla f\)  are \(t\) -related, and \(\nabla ((f\circ s)\circ m)\)  and \(\nabla (f\circ s)\)  are \(m\) -related. Here \(\nabla ((f\circ s)\circ m)\)  denotes the gradient vector field of \((f\circ s)\circ m\)  with respect to \(\eta ^{(2)}\)  on \(G^{(2)}\)  and we have that
\(dm \circ \nabla ((f\circ s)\circ m)=\nabla (f\circ s)\circ m.\) 
The crucial point is to prove that for \((g,h)\in G^{(2)}\)  the identity
\(\nabla ((f\circ s)\circ m)(g,h)=(\nabla (f\circ s)(g),\nabla (f\circ t)(h)),\) 
holds true where \(ds(g)(\nabla (f\circ s)(g))=\nabla f(s(g))=\nabla f(t(h))=dt(h)(\nabla (f\circ t)(h))\) . Take \((v,w)\in TG^{(2)}\)  so that \(ds(g)(v)=dt(h)(w)\) . Thus, by using the identity mentioned in Remark 2.5 from [1]} that relates \(\eta ^{(2)}\) , \(\eta ^{(1)}\)  and \(\eta ^{(0)}\)  we have that
\(& & \eta ^{(2)}_{(g,h)}(\nabla ((f\circ s)\circ m)(g,h)-(\nabla (f\circ s)(g),\nabla (f\circ t)(h)),(v,w))\\& = & d((f\circ s)\circ m)_{(g,h)}(v,w)-\eta ^{(1)}_g(\nabla (f\circ s)(g),v)-\eta ^{(1)}_h(\nabla (f\circ t)(h),w)\\& + & \eta ^{(0)}_{s(g)}(ds(g)(\nabla (f\circ s)(g)),ds(g)(v))\\& = & df(s(gh))(d(s\circ m))_{(g,h)}(v,w))-d(f\circ s)(g)(v)-d(f\circ t)(h)(w)+df(s(g))(ds(g)(v))\\& = & df(s(h))(ds(h)(w))-d(f\circ t)(h)(w)\\& = & d(f\circ s)(h)(w)-d(f\circ t)(h)(w)=0,\) 
since \(s\circ m=s\circ \pi _2\)  and \(s^\ast f=t^\ast f\) . Therefore, as \(\eta ^{(2)}\)  is non-degenerate we get what we required and because of this \(dm \circ (\nabla (f\circ s)\times \nabla (f\circ s))=\nabla (f\circ s)\circ m\) .
